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Topic: Is it possible to bound these functions?
Replies: 3   Last Post: Apr 5, 2013 10:15 PM

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William Elliot

Posts: 1,532
Registered: 1/8/12
Re: Is it possible to bound these functions?
Posted: Apr 5, 2013 10:15 PM
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From marsh@panix.com Fri Apr 5 19:13:04 2013
Date: Fri, 5 Apr 2013 19:13:04 -0700 (PDT)
From: William Elliot <marsh@panix.com>
To: William Elliot <marsh@panix.com>
Subject: Re: Is it possible to bound these functions?

> Define A{f(x)} as a mapping from the set of functions defined on the
interval [0,1] to the Reals.

> The functions are as "nice, smooth and integrable" as you may need them
> to be.
>
> A{f(x)} = [ int cos(int f(t) t=0..x) x=0..1 ]^2 +
> [ int sin(int f(t) t=0..x) x=0..1 ]^2
>

A(f) = (integral(0,1) cos(integral(0,x) f(t) dt) dx)^2
+ (integral(0,1) sin(integral(0,x) f(t) dt) dx)^2

> Given that a <= f(x) <= b, can it be shown that A{a} => A{f(x)} => A{b} ?

No, let a = -2, b = 2, f = 0.
integral(0,x) 0 dx = 0; A(0) = 1
integral(0,1) cos x = sin 1; integral(0,1) sin x dx = cos 1

If a /= 0, then, integral(0,x) a dt = ax;
integal(0,1) cos ax = (sin a)/a; integral(0,1) sin ax dx = (cos a)/a
A(a) = 1/a^2




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